3
$\begingroup$

Does there exist a continuous function of $f : \mathbb R^2 \longrightarrow \mathbb R$ such that it is continuous whose both the partial derivatives don't exist.

I think the function $f : \mathbb R^2 \longrightarrow \mathbb R$ defined by $f(x,y) = |x|(1 + y)$, where $(x,y) \in \mathbb R^2$ has the above property at $(0,0)$. But I can't prove that $f$ is continuous at $(0,0)$ by $\epsilon-\delta$ method. Please help me.

Thank you in advance.

$\endgroup$
  • $\begingroup$ Do you must to prove with the $\;\epsilon-\delta\;$ method? What about $\;||x|(1+y)|\le2|x|\xrightarrow{}0\;$ ? $\endgroup$ – DonAntonio Dec 31 '16 at 17:21
  • $\begingroup$ Do you know that the product of continuous functions is continuous? And that $f(x,y)=|x|$ and $f(x,y)=1+y$ are continuous? It's usually easier to use those properties than $\varepsilon$-$\delta$. $\endgroup$ – Milo Brandt Dec 31 '16 at 17:21
5
$\begingroup$

Just consider the sum of two one-variable Weierstrass functions, one in the $x$ variable, one in the $y$ variable. This is even better, it has continuity everywhere and no differentiability or partial derivatives anywhere.

$\endgroup$
0
$\begingroup$

Let $\epsilon>0$ given.

we have to find $\delta>0$ such that:

$|x-0|<\delta $and $|y-0|<\delta \implies ||x|(1+y)|<\epsilon$.

as we are near $(0,0)$, we can assume that $\color{red}{-0.5}<y<\color{red}{0.5}$. which gives $0.5<1+y<1.5$ and $|1+y|<1.5$.

It is now easier to look for $\delta.\;:$ $|f(x,y)-f(0,0)|=|x||1+y|<1.5|x|<\epsilon$

or $|x|<\frac{2\epsilon}{3}$. thus, we will take $$\delta=\min(\frac{2\epsilon}{3},\color{red}{0.5})$$ to satisfy both conditions

  • $|y|<0.5$

  • $|x|<\frac{2\epsilon}{3}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.